Let us consider a point P having the two-dimensional cartesian co-ordinates (x,y). The polar co-ordinates of that point in the two dimensional system are \( (r,\ \theta) \).

here,

\( x=r\ \cos\theta\tag{1} \)

and \( y=r\ \sin\theta\tag{2} \).

Differentiating equations (1) & (2) with respect to time \( t \) we get,

\( \frac{dx}{dt}=\dot{x}=\dot{r}\ \cos\theta-r\ \sin\theta\ \dot{\theta}\tag{3} \)

and,

\( \dot{y}=\dot{r}\ \sin\theta+r\ \cos\theta\ \dot{\theta}\tag{4} \)

Multiplying equations (3) and (4) by \( cos\theta \) and \( \sin\theta \) respectively and then adding them we get,

\( \dot{x}\ \cos\theta+\dot{y}\ \sin\theta=\dot{r}\left({cos}^2\theta+{\sin}^2\theta\right) \)

\( or,\ \dot{x}\ \cos\theta+\dot{y}\ \sin\theta=\dot{r} \)

\( or,\ \displaystyle{\frac{(\dot{x}\ r\ \cos\theta+\dot{y}\ r\ \sin\theta)}{r}}=\dot{r} \)

\( or,\ \dot{r}=\displaystyle{\frac{\dot{x}x+\dot{y}y}{r}} \)

\( or,\ \dot{r}=\displaystyle{\frac{\dot{x}x+\dot{y}y}{\sqrt{x^2+y^2}}}\tag{5} \)

[ Since, from equations (1) and (2) we get, \( x^2+y^2=r^2{\cos}^2\theta+r^2{\sin}^2\theta=r^2 \) ]

Again multiplying equations (3) and (4) by \sin\theta and\cos\theta respectively and subtracting we get,

\( \displaystyle{\dot{y}\ \cos\theta-\dot{x}\ \sin\theta=r\left(\sin^2\theta+\cos^2\theta\right)\dot{\theta}} \)

\( or,\ \displaystyle{\dot{\theta}=\frac{\dot{y}\ \cos\theta-\dot{x}\ \sin\theta}{r}} \)

\( or,\ \displaystyle{\dot{\theta}=\frac{\dot{y}x-\dot{x}y}{r^2}} \)

\( or,\ \displaystyle{\dot{\theta}=\frac{\dot{y}x-\dot{x}y}{x^2+y^2}} \)

\( or,\ \displaystyle{r\dot{\theta}=\frac{\dot{y}x-\dot{x}y}{r}}\tag{6} \)

Again we know that the radial velocity and transverse velocity components are \( v_r=\dot{r} \) and \( v_{\theta}=r\dot{\theta} \) respectively.

So the Equations (5) and (6) can also be written as,

the radial velocity component is \( \displaystyle{v_{r}=\frac{\dot{x}x+\dot{y}y}{r}} \)

and the transverse velocity component is \( \displaystyle{v_{\theta}= \frac{\dot{y}x-\dot{x}y}{r}} \)